# Homework Help: Find the General Anti-derivative (Calculus I)

1. Apr 29, 2012

### treehau5

1. The problem statement, all variables and given/known data
given f(x) = [x^3+5sqrt(x)]/x^2, find the anti-derivative

2. Relevant equations

3. The attempt at a solution

Hi I have attempted to solve this by re-writing the equation as a sum of two fractions:

x^3/x^2 + 5sqrt(x)/x^2, simplying gives = x + 5/x^3/2

I then apply the principle anti-differentiation rules, and I come out with:

x^2/2 + 5x^-1/2 = x^2/2 + 10/sqrt(x) + C

This answer is coming back in WebWork as wrong. I also checked my answer on wolframalpha, and got the same result.

What am I doing wrong? (If anything)

2. Apr 29, 2012

### sharks

Try re-writing it: $$x + \frac{5}{x^{3/2}}=x+5x^{-3/2}$$Then integrate.

3. Apr 29, 2012

### treehau5

I am not quite at intergration yet, this is the last section of my Calculus I course, Calculus II is intergration. So for now our anti-d's are pretty simple and just follow some basic rules.

4. Apr 29, 2012

### treehau5

Ok so
x + 5x-3/2 following the rule: x(n+1)/(n+1) gives

x1+1 / (1+1) + 5x-3/2 + 2/2 / -(1/2) =

x2/2 + 5x-1/2/ -(1/2 ) = -10x-1/2 or finally

x2 / 2 + 10 / sqrt(x)

I am still doing it wrong?

5. Apr 29, 2012

### treehau5

ok yes I am doing it wrong, its - 10 / sqrt(x).

Thank you very much.

Story of my life, I always miss a sign.

6. Apr 29, 2012

### Staff: Mentor

Integration is finding the antiderivative.

Note that there is no such word in English as intergration.

7. Apr 29, 2012

### treehau5

Thank you Mark for the insight. And you are right, there is no such word as intergration.